How to check if a number is odd?

[rpkamp]

Based on this:

The idea here that this is obviously wrong because there are better ways to do this. So, what ways are there?

Disclaimer: This topic is just meant as discussion, I know several answers, just wondering what others come up with.

[James_Hibbard]

Easiest way would be to use the modulus operator:

const isEven = num => num%2 === 0;

isEven(199)
// false

isEven(200)
// true

[DaveMaxwell]

What he said. Time tested method. Divide by two and if there’s a remainder, it’s odd.

[kerry14]

yep, I agree modulo - in php…
fmod(x,y) returns 0 if no remainder
so divide by 2
so fmod(10, 2) returns 0
whereas
fmod(7, 2) returns 1

and since it’s just a discussion I guess any string resulting from a division by 2, if the result was even would only contain 0s after the ‘.’ or no ‘.’ at all

however I am not sure what happens with 0 as one of the arguments - is 0 an odd number or even ?

Just for fun, apparently if it is a whole number (not a decimal) if the last digit is 0,2,4,6 or 8 then it is even, any other digit and it’s odd

so even without a calculator …
12345678901234567890 is even, but
12345678901234567891 is odd
just check the last digit !

[tracknut]

You might want to define “even” and “odd” as, at least on the internet, there seems to be some debate.

[rpkamp]

how can there be debate about this? they are clearly defined mathematical concepts

Let’s go with

even = integer that can be divided exactly by 2
odd = integer that can not be divided exactly by 2

[tracknut]

Okay, so integers only, positive and negative allowed, how about this in php:

function isEven ($n)
	{
	if (gettype ($n) != "integer") return false;
	return (($n & 1) == 0) ? 1 : 0;
	}

Edit: I don’t know why my brain turned the question into checking if a number is even rather than odd, but the switch is… apparent.

[James_Hibbard]

Using the venerable isOdd package:

const isOdd = require('is-odd');
 
console.log(isOdd('1')); //=> true
console.log(isOdd('3')); //=> true
 
console.log(isOdd(0)); //=> false
console.log(isOdd(2)); //=> false

Before you knock it, bear in mind that it has 483k weekly downloads…

Here is how the experts do it: https://github.com/jonschlinkert/is-odd/blob/master/index.js

[SamA74]

In PHP.

if($n & 1){}

If it’s true, it’s odd.

[annabrone24]

In python you can write it as
Odd=numbers which are not divisible by 2

Program Solution

x=input("enter any number")
if(x/2!=0);
  print("number is odd")
else
  print("number is even")

[rpkamp]

Really inefficient, but works too :

function isOdd($n): bool {
    while ($i >= 0) {
        $i -= 2;
    }
    return $i === -1;
}

[benanamen]

Fun with Php

<?php

$arr = range(3,9000,2);
$num = 9;

if (in_array($num, $arr)){
    echo "Num is odd";
}

[rpkamp]

Oh, oh, generators!

<?php
function isOdd($i) {
    if ($i === 0) { return false; }

    $sign = $i > 0 ? 1 : -1;

    $oddNumberGenerator = function() use ($sign) {
        $i = $sign; // start at -1 when $i is positive, at 1 when $i is negative
        while (true) {
            $i += $sign * 2; // add 2 when $i is positive, substract 2 when $i is negative
            yield $i;
        }
    };

    foreach ($oddNumberGenerator() as $oddNumber) {
        if ($i === $oddNumber) { return true; }
        if ($i === $oddNumber + 1) { return false; }
    }
}

It’s horribly ineffecient. Just deciding whether 99999999 is odd takes 10 seconds on https://repl.it/languages/php

[benanamen]

Wait, wait, how bout this?

<?php

$num =216259;
$x = substr($num, -1);

if  ($x == 1 || $x == 3 ||$x == 5 || $x == 9){
    echo 'Num is odd';
}

[rpkamp]

Or this

function isOdd($i) {
    return substr((string) ($i / 2), -2) === '.5';
}

[lenmerkel]

You didn’t specify which programming language. In many languages, the likely most efficient method would be to use a bitwse operator. For example, in Javascript, like so:

const isOdd = value & 1

Here we’re ANDing the value with 1, which will return 1 (truthy) if and only if the right-hand bit of value is also 1. Odd values have that right-hand bit set while even values (and zero) don’t.

[m3g4p0p]

And then of course using recursion:

function isOdd (value) {
  return value !== 0 && (value === -1 || isOdd(Math.abs(value) - 2))
}